audioguru

When the battery voltage is low or when the battery is not connected to the collectors of the output transistors then when the BD139 turns on its load is the base-emitter diodes of the output transistor in series with the very low value resistors causing a very high current in the BD139. I said yesterday to add a resistor between the emitter of the BD139 and the bases of the output resistors to limit the current.

I see why the BD139 draws a high current in your battery discharger circuit: There should be a current-limiting resistor between the emitter of the BD139 and the bases of the output transistors.

Opamp U2, the BD139 emitter-follower and the output transistors emitter-followers make a DC amplifier with a voltage gain of about 3.07 times. If the input to U2 is 0V then the output of U2 is about +1V and the output of the project is 0V. If the input to U2 is +5V then its output is about +6.5V and the output of the project is(5V x 3.07=) +15.35V. if the input to U2 is 9.77V then the output is about +11.3V and the output of the project is +29.99V. Oh, maybe you are talking about the battery discharger circuit we discussed a few years ago? I do not think its schematics are here any more.

The current in a telephone line is no where near 1A, if you are close to the exchange it is about 50mA and if you are far it is maybe 10mA. The power that feeds a telephone has a 300 ohm resistor (a relay coil) on each of the two wires so the phone transmits into 600 ohms and receives from 600 ohms. Two telephones will work from a 12V battery if you connect 300 ohms to the +12V and another 300 ohms to the 0V. Then connect both phones to the free ends of the two resistors.

Like I said, maybe an output transistor is shorted (or wired wrong) making the output voltage of the power supply as high as it can go. Since you removed the BD139 transistor then the diode D10 on the original schematic connected to it and to the project's output conducts and forces the output of the opamp to a high voltage and maybe destroying the opamp.

The opamp , BD139 or an output transistor is shorted.

Kevin, you do not make any sense.

If you have 0V on the (+) input pin of the LM324 opamp and its (-) input is at a positive voltage of about 0.6V then its output should be as low as it can go which is about 0.01V so thall the transistors should be turned off. If the voltage at the 0.5 ohm resistor is 0.6V then the bases of the output transistors should be about 0.86V and the base of the BD139 should be about 1.6V. Then the input should be 0.6V. When the input voltage is less then the voltage at the 0.5 ohm resistor should be the same.

There is no voltage multiplier in this thread.

Using the EDGE browser on Windows 10 when I clicked on "reply" it filled in the text I already posted a week or two ago. When I changed the browser to Internet Explorer then I needed to log in (again) and go through a few pages to find this thread. I will try to post an attachment, but I can't since the box that I must click on is not showing and when I hover my mouse pointer over where the box should be then nothing happens.

Thanks for the freelancing job recommendation but I am a 70 years old government worker (I am retired with a government pension and some stocks and bonds) so I do not need more income. I do whatever I want whenever I want.

Distortion can blow out a tweeter because distortion has strong high frequency harmonics. Try feeding an input directly to the power amplifier without connecting the preamp.

The output of a preamp will produce clipping distortion if the input level is too high. The pickup from an electric guitar produces a level much higher than an FM tuner, a CD player or an MP3 player. Most preamps for a guitar pickup use a Jfet because of its very high input impedance with a gain of only about 1.4 times.

The 25VAC transformer produces a peak voltage of 35.4V which is reduced to 33.4V by the rectifier bridge. Then try to turn the voltage down to 1.2V at 1.5A then the regulator will heat with (33.4V - 1.2V) x 1.5A= 48.3W which is way too much heat for it. Luckily it has thermal protection built-in so it shuts down when it gets too hot. If the output is set to 5V and loaded with only 1A then the LM317 will heat with (33.4V - 5V) x 1A= 28.4W which will also cause it to shut down even if it has a HUGE heatsink. Therefore reduce the voltage from the transformer.

Nobody makes an LA4058. You got the numbers mixed up, the schematic shows a Sanyo LA4508. Your 13.6V transformer must feed the rectifiers on the Power Part of the schematic. Maybe you fed the transformer wrongly to the amplifier instead of to the rectifiers? Then probably many parts are destroyed. Your 13.6V transformer will produce a peak of 19.2V and the rectifiers reduce it to +17.2V. With a 17V supply the LA4508 produces about 6W at clipping into 4 ohms per channel. Who told you 100W? It has fairly high distortion at low and high frequencies and it cuts frequencies above only 5kHz. Your amplifier probably uses one LA4508 for left woofer and left tweeter and the second LA4508 is probably used for right woofer and right tweeter.

I cannot make an attachment here with this new software when I had Windows 8.1 and Internet Explorer 11 and now I also cannot on Windows 10 and Edge or with Internet Explorer 11 even when I activated the latest JavaScript that I never needed to use on any other site. Maybe this new software works only from a Macintosh operating system?

The same current flows through parts that are in series so their sequence does not matter. If your series circuits have the same parts values and the same AC input signal then their capacitors charge exactly the same. Your circuits have no input signal so nothing gets halved. A diode rectifies, it does not halve.

C9 is supposed to be 100pf, not 10pF.

The TDA7297 is not small, it has a normal size for an audio amplifier IC. Its datasheet has a recommended pcb design.

Maybe an electronic high voltage generator circuit can zap the driver when he tries to adjust the drivers seat. Maybe a solenoid can be activated to lock the seat position when the car is started.

The browser called EDGE in Windows 10 does not support plugins so I installed Java on Internet Explorer that I am now using. I still do not have a button for attachments. Anti-virus is turned on. I added Java, re-booted then checked that Java is active. I changed from Browser Edge to browser Internet Explorer (a nuisance) so that Java is running. I disabled AVG anti-virus and I STILL CANNOT ATTACH ANYTHING. Is there anything else I can do? Other electronics forums work perfectly.

I disabled my AVG anti-virus program and I still do not see an attachments button.

In Windows 10 I turned off its popup blocker and turned off its Smartscreen Filter. Then I re-booted and checked that they are still turned off. I still do not see the attachments button. I looked at how to add JavaScript but it said to do many things I do not know how to do.

This time I changed Windows 10 to Internet Explorer mode then I needed to log in here again. I still do not see the button for attachments. I also looked at Internet Explorer addons and millions of advanced settings but I did not see a javascript plugin. I do not know how to turn off my AVG Free anti-virus program.

This thread has been for such a long time that most schematics are gone. I do not know what you want to do. A charger puts current into a battery from a voltage higher than the battery voltage. The current is limited or regulated. A battery is discharged by a load that draws current out of the battery. The load current is limited.